carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o).\nco(g) + 3h₂(g) ⇌…

carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o).\nco(g) + 3h₂(g) ⇌ ch₄(g) + h₂o(g)\nthe reaction is at equilibrium at 1,000 k. the equilibrium constant of the reaction is 3.90. at equilibrium, the concentrations are as follows.\nco = 0.30 m\nh₂ = 0.10 m\nh₂o = 0.020 m\nwhat is the equilibrium concentration of ch₄ expressed in scientific notation?\n0.0059\n5.9×10⁻²\n0.059\n5.9×10²

carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o).\nco(g) + 3h₂(g) ⇌ ch₄(g) + h₂o(g)\nthe reaction is at equilibrium at 1,000 k. the equilibrium constant of the reaction is 3.90. at equilibrium, the concentrations are as follows.\nco = 0.30 m\nh₂ = 0.10 m\nh₂o = 0.020 m\nwhat is the equilibrium concentration of ch₄ expressed in scientific notation?\n0.0059\n5.9×10⁻²\n0.059\n5.9×10²

Answer

Answer:

$5.9\times 10^{-2}$

Explanation:

Step1: Write equilibrium - constant expression

$K = \frac{[CH_4][H_2O]}{[CO][H_2]^3}$

Step2: Rearrange for $[CH_4]$

$[CH_4]=\frac{K[CO][H_2]^3}{[H_2O]}$

Step3: Substitute given values

$[CH_4]=\frac{3.90\times0.30\times(0.10)^3}{0.020}$

Step4: Calculate result

$[CH_4] = 0.0585\approx5.9\times 10^{-2}$